Integration by substitution

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A triangle with sides a, b, and c.

In geometry, Heron's (or Hero's) formula, named after Hero of Alexandria, states that the area T of a triangle whose sides have lengths a, b, and c is

$T={\sqrt {s(s-a)(s-b)(s-c)}}$

where s is the semiperimeter of the triangle:

$s={\frac {a+b+c}{2}}.$

Heron's formula can also be written as:

$T={\frac {1}{4}}{\sqrt {(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}$

$T={\frac {1}{4}}{\sqrt {2(a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2})-(a^{4}+b^{4}+c^{4})}}$

$T={\frac {1}{4}}{\sqrt {(a^{2}+b^{2}+c^{2})^{2}-2(a^{4}+b^{4}+c^{4})}}$

$T={\frac {1}{4}}{\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}$

Heron's formula is distinguished from other formulas for the area of a triangle, such as half the base times the height or half the modulus of a cross product of two sides, by requiring no arbitrary choice of side as base or vertex as origin.

Example

Let ΔABC be the triangle with sides a=7, b=4 and c=5. The semiperimeter is   $s={\tfrac {1}{2}}(a+b+c)={\tfrac {1}{2}}(7+4+5)=8$ , and the area is

 $T={\sqrt {s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}={\sqrt {8\cdot (8-7)\cdot (8-4)\cdot (8-5)}}$

$={\sqrt {8\cdot 1\cdot 4\cdot 3}}={\sqrt {96}}=4{\sqrt {6}}\approx 9.8$

History

The formula is credited to Heron (or Hero) of Alexandria, and a proof can be found in his book, Metrica, written c. A.D. 60. It has been suggested that Archimedes knew the formula over two centuries earlier, and since Metrica is a collection of the mathematical knowledge available in the ancient world, it is possible that the formula predates the reference given in that work.

A formula equivalent to Heron's namely:

$T={\frac {1}{2}}{\sqrt {a^{2}c^{2}-\left({\frac {a^{2}+c^{2}-b^{2}}{2}}\right)^{2}}}$, where $a\geq b\geq c$

was discovered by the Chinese independently of the Greeks. It was published in Shushu Jiuzhang (“Mathematical Treatise in Nine Sections”), written by Qin Jiushao and published in A.D. 1247.

Proof

A modern proof, which uses algebra and is quite unlike the one provided by Heron (in his book Metrica), follows. Let a, b, c be the sides of the triangle and A, B, C the angles opposite those sides. We have

$\cos {\widehat {C}}={\frac {a^{2}+b^{2}-c^{2}}{2ab}}$

by the law of cosines. From this proof get the algebraic statement:

$\sin {\widehat {C}}={\sqrt {1-\cos ^{2}{\widehat {C}}}}={\frac {\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}{2ab}}.$

The altitude of the triangle on base a has length b·sin(C), and it follows

${\begin{aligned}T&={\frac {1}{2}}({\mbox{base}})({\mbox{altitude}})\\&={\frac {1}{2}}ab\sin {\widehat {C}}\\&={\frac {1}{4}}{\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}\\&={\frac {1}{4}}{\sqrt {(2ab-(a^{2}+b^{2}-c^{2}))(2ab+(a^{2}+b^{2}-c^{2}))}}\\&={\frac {1}{4}}{\sqrt {(c^{2}-(a-b)^{2})((a+b)^{2}-c^{2})}}\\&={\sqrt {\frac {(c-(a-b))(c+(a-b))((a+b)-c)((a+b)+c)}{16}}}\\&={\sqrt {{\frac {(b+c-a)}{2}}{\frac {(a+c-b)}{2}}{\frac {(a+b-c)}{2}}{\frac {(a+b+c)}{2}}}}\\&={\sqrt {{\frac {(a+b+c)}{2}}{\frac {(b+c-a)}{2}}{\frac {(a+c-b)}{2}}{\frac {(a+b-c)}{2}}}}\\&={\sqrt {s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}.\end{aligned}}$

The difference of two squares factorization was used in two different steps.

Proof using the Pythagorean theorem

Triangle with altitude h cutting base c into d + (c − d).

Heron's original proof made use of cyclic quadrilaterals, while other arguments appeal to trigonometry as above, or to the incenter and one excircle of the triangle [2]. The following argument reduces Heron's formula directly to the Pythagorean theorem using only elementary means.

We wish to prove $4T^{2}=4s(s-a)(s-b)(s-c).$ The left-hand side equals

$4T^{2}=(ch)^{2}=c^{2}(b^{2}-d^{2})=(cb)^{2}-(cd)^{2}$

while the right-hand side equals

$4s(s-a)(s-b)(s-c)=[s(s-a)+(s-b)(s-c)]^{2}-[s(s-a)-(s-b)(s-c)]^{2}$

via the identity $(p+q)^{2}-(p-q)^{2}=4pq.$ It therefore suffices to show

$cb=s(s-a)+(s-b)(s-c)$

and

$cd=s(s-a)-(s-b)(s-c).$

Substituting $2s=(a+b+c)$ into the former,

$s(s-a)+(s-b)(s-c)={\frac {1}{4}}(a+b+c)(-a+b+c)+{\frac {1}{4}}(a-b+c)(a+b-c)={\frac {1}{4}}[(b+c)^{2}-a^{2}]+{\frac {1}{4}}[a^{2}-(b-c)^{2}]={\frac {1}{4}}[(b+c)^{2}-(b-c)^{2}]=cb$

as desired. Similarly, the latter expression becomes

$s(s-a)-(s-b)(s-c)={\frac {1}{4}}[(b+c)^{2}-a^{2}]-{\frac {1}{4}}[a^{2}-(b-c)^{2}]={\frac {1}{2}}(b^{2}+c^{2}-a^{2}).$

Using the Pythagorean theorem twice, $b^{2}=d^{2}+h^{2}$ and $a^{2}=(c-d)^{2}+h^{2},$ allows us to simplify the expression to

${\frac {1}{2}}(b^{2}+c^{2}-a^{2})={\frac {1}{2}}[d^{2}+c^{2}-(c-d)^{2}]=cd.$

The result follows.

Proof using the Law of cotangents and the triple cotangent identity

Geometrical significance of s-a, s-b, and s-c. See the Law of cotangents for the reasoning behind this.

From the first part of the Law of cotangents proof, we have that the triangle's area is both

${\begin{aligned}T&=r{\big (}(s-a)+(s-b)+(s-c){\big )}=r^{2}\left({\frac {s-a}{r}}+{\frac {s-b}{r}}+{\frac {s-c}{r}}\right)\\[8pt]&=r^{2}{\big (}\cot(A/2)+\cot(B/2)+\cot(C/2){\big )}\\[8pt]\end{aligned}}$

and

$T=rs$

but, since the sum of the half-angles is ${\tfrac {\pi }{2}}$, the triple cotangent identity applies, so the first of these is

${\begin{aligned}T&=r^{2}\left(\cot(A/2)\ \cot(B/2)\ \cot(C/2)\right)=r^{2}\ {\frac {s-a}{r}}\ {\frac {s-b}{r}}\ {\frac {s-c}{r}}\\[8pt]&={\frac {(s-a)(s-b)(s-c)}{r}}\\[8pt]\end{aligned}}$

Combining the two, we get

$T^{2}=s(s-a)(s-b)(s-c)$

from which the result follows.

Numerical stability

Heron's formula as given above is numerically unstable for triangles with a very small angle. A stable alternative involves arranging the lengths of the sides so that $a\geq b\geq c$ and computing

$T={\frac {1}{4}}{\sqrt {(a+(b+c))(c-(a-b))(c+(a-b))(a+(b-c))}}.$

The brackets in the above formula are required in order to prevent numerical instability in the evaluation.

Other area formulas resembling Heron's formula

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Several other triangle area formulas have the same functional form as Heron's formula but with the sides replaced by either the medians, the reciprocals of the altitudes, or the sines of the angles.

Generalizations

Heron's formula is a special case of Brahmagupta's formula for the area of a cyclic quadrilateral. Heron's formula and Brahmagupta's formula are both special cases of Bretschneider's formula for the area of a quadrilateral. Heron's formula can be obtained from Brahmagupta's formula or Bretschneider's formula by setting one of the sides of the quadrilateral to zero.

Heron's formula is also a special case of the formula for the area of a trapezoid or trapezium based only on its sides. Heron's formula is obtained by setting the smaller parallel side to zero.

Expressing Heron's formula with a Cayley–Menger determinant in terms of the squares of the distances between the three given vertices,

$T={\frac {1}{4}}{\sqrt {-{\begin{vmatrix}0&a^{2}&b^{2}&1\\a^{2}&0&c^{2}&1\\b^{2}&c^{2}&0&1\\1&1&1&0\end{vmatrix}}}}$

illustrates its similarity to Tartaglia's formula for the volume of a three-simplex.

Another generalization of Heron's formula to pentagons and hexagons inscribed in a circle was discovered by David P. Robbins.

Heron-type formula for the volume of a tetrahedron

If U, V, W, u, v, w are lengths of edges of the tetrahedron (first three form a triangle; u opposite to U and so on), then

${\text{volume}}={\frac {\sqrt {\,(-a+b+c+d)\,(a-b+c+d)\,(a+b-c+d)\,(a+b+c-d)}}{192\,u\,v\,w}}$

where

${\begin{aligned}a&={\sqrt {xYZ}}\\b&={\sqrt {yZX}}\\c&={\sqrt {zXY}}\\d&={\sqrt {xyz}}\\X&=(w-U+v)\,(U+v+w)\\x&=(U-v+w)\,(v-w+U)\\Y&=(u-V+w)\,(V+w+u)\\y&=(V-w+u)\,(w-u+V)\\Z&=(v-W+u)\,(W+u+v)\\z&=(W-u+v)\,(u-v+W).\end{aligned}}$

See also

• Heronian triangle

References

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External links

• A Proof of the Pythagorean Theorem From Heron's Formula at cut-the-knot
• Interactive applet and area calculator using Heron's Formula
• J.H. Conway discussion on Heron's Formula
• Template:MathPages
• A Geometric Proof of Heron's Formula